Three boys are pushing horizontally a box placed on horizontal table. One is pushing towards north with a `5sqrt(3)`N force. The second is pushing towards east and third pushes with a force 10 N such that the box is in equilibrium. Find the magnitude of the force, second by is applying in newton.

Answer

To solve the problem step by step, we will analyze the forces acting on the box and use the equilibrium condition to find the force applied by the second boy.

### Step 1: Identify the Forces
- The first boy applies a force \( F_1 = 5\sqrt{3} \, \text{N} \) towards the north.
- The second boy applies a force \( F_2 = x \, \text{N} \) towards the east (we need to find \( x \)).
- The third boy applies a force \( F_3 = 10 \, \text{N} \) in a direction opposite to the resultant of the first two forces to maintain equilibrium.

### Step 2: Calculate the Resultant Force
The resultant force \( R \) of the first two forces can be calculated using the Pythagorean theorem since they are perpendicular to each other:

\[ R = \sqrt{F_1^2 + F_2^2} \]

Substituting \( F_1 \) and \( F_2 \):

\[ R = \sqrt{(5\sqrt{3})^2 + x^2} \]

Calculating \( (5\sqrt{3})^2 \):

\[ (5\sqrt{3})^2 = 25 \times 3 = 75 \]

So, we have:

\[ R = \sqrt{75 + x^2} \]

### Step 3: Apply the Equilibrium Condition
For the box to be in equilibrium, the resultant force \( R \) must equal the force applied by the third boy:

\[ R = F_3 = 10 \, \text{N} \]

Thus, we can set up the equation:

\[ \sqrt{75 + x^2} = 10 \]

### Step 4: Square Both Sides
To eliminate the square root, we square both sides:

\[ 75 + x^2 = 100 \]

### Step 5: Solve for \( x^2 \)
Rearranging the equation gives:

\[ x^2 = 100 - 75 \]
\[ x^2 = 25 \]

### Step 6: Find \( x \)
Taking the square root of both sides:

\[ x = \sqrt{25} = 5 \]

### Conclusion
The magnitude of the force applied by the second boy is:

\[ \boxed{5 \, \text{N}} \]